sums of floor and sqrt functions

Let $n\in\mathbb{N}$ fixed. How do you compute $$\sum_{n^2<m<(n+1)^2}\left\lfloor\frac{1}{\sqrt{m}-\lfloor\sqrt{m}\rfloor}\right\rfloor?$$

I did some examples for values of $n$, but didn't find a pattern.

Write $m=n+k$ and rationalize the denominator to give \begin {align}\sum_{n^2<m<(n+1)^2}\left\lfloor\frac{1}{\sqrt{m}-\lfloor\sqrt{m}\rfloor}\right\rfloor&=\sum_{k=1}^{2n}\left\lfloor\frac{1}{\sqrt{n+k}-n}\right\rfloor\\&=\sum_{k=1}^{2n}\left\lfloor\frac{\sqrt{n+k}+n}{k}\right\rfloor\\&=\sum_{k=1}^{2n}\left\lfloor\frac{2n}{k}\right\rfloor \end {align} showing that as Hw Chu comments this is the even numbered terms of OEIS A006218.